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The tuba typically produces the lowest-frequency sounds in an orchestra. Consider a tuba to be a coiled tube of length \(7.373 \mathrm{~m}\). What is the lowest-frequency sound a tuba can produce? Assume that the speed of sound is \(343.0 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
Answer: The lowest frequency sound a tuba can produce is approximately 23.3 Hz.

Step by step solution

01

Open pipe fundamentals

We have a coiled tube, and we can consider the tube as an open pipe because both ends are open. The formula for the fundamental frequency (lowest frequency) of an open pipe is: \(f_1 = \frac{v}{2L}\) Where: \(f_1\) - the fundamental frequency (Hz) \(v\) - the speed of sound in the air (m/s) \(L\) - the length of the pipe (m) In this case, the length of the tuba, \(L = 7.373 \mathrm{~m}\), and the speed of sound, \(v=343.0 \mathrm{~m} / \mathrm{s}\).
02

Calculate the fundamental frequency

Now, we substitute the given values into the formula to find the lowest frequency: \(f_1 = \frac{343.0 \mathrm{~m} / \mathrm{s}}{2 \times 7.373 \mathrm{~m}}\) Calculate the denominator first: \(2 \times 7.373 \mathrm{~m} = 14.746 \mathrm{~m}\) Now, divide the numerator by the denominator: \(f_1 = \frac{343.0}{14.746} = 23.251 \mathrm{~Hz}\)
03

Round the result

The result should be rounded to an appropriate number of significant figures. In this case, we have three significant figures in the given values, so the result should also have three significant figures: Lowest-frequency sound a tuba can produce: \(f_1 \approx 23.3 \mathrm{~Hz}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acoustic Resonance
Acoustic resonance is a phenomenon where an object vibrates in response to external sound waves whose frequency matches the object's natural frequency. Imagine blowing across the top of an empty bottle and hearing a tone. That tone is a result of acoustic resonance, where the air inside the bottle vibrates at a particular frequency.

In musical terms, the tuba, like other wind instruments, makes use of this principle. The air column inside its long, coiled pipe vibrates sympathetically with the player's buzzing lips on the mouthpiece. The wave reflects back and forth within the pipe, and when it constructively interferes with itself at certain frequencies, resonance occurs, producing a rich sound.
Harmonics in Musical Instruments
Harmonics are integral to music, as they contribute to the richness and timbre of the sound produced by musical instruments. Each fundamental frequency, or the lowest frequency at which an instrument can play a note, is accompanied by higher frequencies called overtones or harmonics. These harmonics are integer multiples of the fundamental frequency.

Instruments like the tuba produce a fundamental tone, but the sound we hear contains a mix of many frequencies. The player can modify the pitch by changing the lip tension and by altering the length of the air column, using valves on the instrument, to produce these harmonics.
Speed of Sound
The speed of sound in air, approximately 343 m/s at 20°C, is fundamental to understanding how wind instruments function. The speed at which sound waves travel through a medium like air can be affected by temperature, humidity, and pressure. However, when calculating the fundamental frequency of an instrument such as the tuba, we use a standard value for the speed of sound to maintain uniformity.

It's this speed that determines how quickly the sound waves produced by the instrument reach our ears and the pitch of the sounds that the instrument generates.
Wavelength and Frequency Relationship
The relationship between wavelength and frequency is pivotal when analyzing sounds produced by musical instruments. This relationship is inversely proportional, as represented by the equation:
\[ v = f \lambda \] Where:\( v \) represents the speed of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength of the sound wave. When you increase the frequency of a wave (making the pitch higher), the wavelength becomes shorter, and vice versa.

This relationship is why the tuba, with its long pipe, naturally produces low-frequency sounds. The longer length of the pipe allows for longer wavelengths, which correspond to lower frequencies. Critically, understanding this relationship helps in grasively on the pitch of sounds produced.

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Most popular questions from this chapter

Consider a sound wave (that is, a longitudinal displacement wave) in an elastic medium with Young's modulus \(Y\) (solid) or bulk modulus \(B\) (fluid) and unperturbed density \(\rho_{0}\). Suppose this wave is described by the wave function \(\delta x(x, t),\) where \(\delta x\) denotes the displacement of a point in the medium from its equilibrium position, \(x\) is position along the path of the wave at equilibrium, and \(t\) is time. The wave can also be regarded as a pressure wave, described by wave function \(\delta p(x, t),\) where \(\delta p\) denotes the change of pressure in the medium from its equilibrium value. a) Find the relationship between \(\delta p(x, t)\) and \(\delta x(x, t),\) in general. b) If the displacement wave is a pure sinusoidal function, with amplitude \(A\), wave number \(\kappa,\) and angular frequency \(\omega,\) given by \(\delta x(x, t)=A \cos (\kappa x-\omega t)\) what is the corresponding pressure wave function, \(\delta p(x, t) ?\) What is the amplitude of the pressure wave?

Two vehicles carrying speakers that produce a tone of frequency \(1000.0 \mathrm{~Hz}\) are moving directly toward each other. Vehicle \(\mathrm{A}\) is moving at \(10.00 \mathrm{~m} / \mathrm{s}\) and vehicle \(\mathrm{B}\) is moving at \(20.00 \mathrm{~m} / \mathrm{s}\). Assume that the speed of sound in air is \(343.0 \mathrm{~m} / \mathrm{s},\) and find the frequencies that the driver of each vehicle hears.

A sound level of 50 decibels is a) 2.5 times as intense as a sound of 20 decibels. b) 6.25 times as intense as a sound of 20 decibels. c) 10 times as intense as a sound of 20 decibels. d) 100 times as intense as a sound of 20 decibels. e) 1000 times as intense as a sound of 20 decibels.

Find the resonance frequency of the ear canal. Treat it as a half-open pipe of diameter \(8.0 \mathrm{~mm}\) and length \(25 \mathrm{~mm}\). Assume that the temperature inside the ear canal is body temperature \(\left(37^{\circ} \mathrm{C}\right)\).

You are playing a note that has a fundamental frequency of \(400 .\) Hz on a guitar string of length \(50.0 \mathrm{~cm} .\) At the same time, your friend plays a fundamental note on an open organ pipe, and 4 beats per second are heard. The mass per unit length of the string is \(2.00 \mathrm{~g} / \mathrm{m} .\) Assume that the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). a) What are the possible frequencies of the open organ pipe? b) When the guitar string is tightened, the beat frequency decreases. Find the original tension in the string. c) What is the length of the organ pipe?

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